Question

a particle is projected at 60 degrees with the horizontal with a kinetic energy K.What is the kinetic energy at its highest point ?

Answer 1

let u be the initial velocity.
∴ initial kinetic energy is K = (1/2)mu²
at the maximum height velocity is u cos60 = (u/2)
∴ kinetic energy at the maximum height is KE=(1/2)m(u/2)²
                                                                        =(1/4)k

Answer 2

Answer》》》k/4

Explanation》》》

Kinetic energy at highest point,

(KE)H = 1/2 mv^2 cos^2 θ

= K cos^2 θ

= K(cos 60°)^2

= K/4 .........(Ans.)